Problem: $g(x) = x^{2}+2x+3(f(x))$ $h(t) = -5t^{2}+4(f(t))$ $f(n) = -n^{2}$ $ g(h(0)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(0)$ . Then we'll know what to plug into the outer function. $h(0) = -5(0^{2})+4(f(0))$ To solve for the value of $h$ , we need to solve for the value of $f(0)$ $f(0) = -0^{2}$ $f(0) = 0$ That means $h(0) = -5(0^{2})+(4)(0)$ $h(0) = 0$ Now we know that $h(0) = 0$ . Let's solve for $g(h(0))$ , which is $g(0)$ $g(0) = 0^{2}+(2)(0)+3(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = -0^{2}$ $f(0) = 0$ That means $g(0) = 0^{2}+(2)(0)+(3)(0)$ $g(0) = 0$